Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int x[10004], y[10004];
vector<pair<int, int> > xans, yans;
void add(vector<pair<int, int> > &ans, int a, int b) {
  ans.push_back(make_pair(a + 1, b + 1));
}
void basis(int x[], vector<pair<int, int> > &ans, int n) {
  int i, j, size = 0, k;
  for (i = (1 << 30); i > 0; i >>= 1) {
    for (j = size; j < n; j++)
      if (x[j] & i) break;
    if (j != n) {
      if (j != size) {
        swap(x[j], x[size]);
        add(ans, j, size);
        add(ans, size, j);
        add(ans, j, size);
      }
      for (k = 0; k < n; k++)
        if (k != size && (x[k] & i)) x[k] ^= x[size], add(ans, k, size);
      size++;
    }
  }
}
void print(int a[], int n) {
  int i;
  for (i = 0; i < n; i++) printf("%d ", a[i]);
  printf("\n");
}
char MSB(int num) {
  char ret = 0, i = 1;
  while (num) {
    if (num % 2) ret = i;
    num = num / 2;
    i++;
  }
  return ret;
}
int main() {
  int n, i, idx, j;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &x[i]);
  for (i = 0; i < n; i++) scanf("%d", &y[i]);
  basis(x, xans, n);
  basis(y, yans, n);
  bool flag = true;
  int xor_val;
  for (i = 0; i < n && flag; i++) {
    if (!y[i]) {
      x[i] ^= x[i];
      add(xans, i, i);
      continue;
    }
    idx = -1;
    xor_val = 0;
    for (j = 0; j < n; j++) {
      if (MSB(y[i] & x[j]) == MSB(x[j])) xor_val = xor_val ^ x[j];
      if (MSB(y[i] & x[j]) == MSB(x[j]) && MSB(y[i] & x[j]) == MSB(y[i]))
        idx = j;
    }
    if (xor_val != y[i] || idx == -1) {
      flag = false;
      break;
    }
    if (i != idx) {
      swap(x[i], x[idx]);
      add(xans, i, idx);
      add(xans, idx, i);
      add(xans, i, idx);
    }
    for (j = 0; j < n; j++) {
      if (MSB(y[i] & x[j]) == MSB(x[j]) && j != i) {
        add(xans, i, j);
        x[i] ^= x[j];
      }
    }
  }
  if (!flag)
    printf("-1\n");
  else {
    printf("%d\n", xans.size() + yans.size());
    for (i = 0; i < xans.size(); i++)
      printf("%d %d\n", xans[i].first, xans[i].second);
    for (i = yans.size() - 1; i >= 0; i--)
      printf("%d %d\n", yans[i].first, yans[i].second);
  }
  return 0;
}